Functional equations-doubt

yes, sir....i guessed the ans...but i dont know the proof

eureka... in the problem, it has not been specified whether f is differentiable.. it only states continuity.. so your solution above is not exactly valid.

find f(2),f(3)......f(n) and u can the generalise the series.

It is easy to see that we have
f(x₁ + x₂ + x₃ + . . . + x_n) = f(x₁) + f(x₂) + f(x₃)+ . . . + f(x_n)
Setting all the x_i's equal to 1, we get
f(n)=n f(1)
So taking k = f(1), we see that for all naturals n, we have
f(n)=kn ----- (1)
Setting y = -x in the given functional equation, we get f(0)=f(x)+f(-x)
But f(0) =0 (which can be obtained by setting x=y=0), hence f(-x)=-f(x) implying that f is odd.
Accordingly, if x be a negative integer, set x = -n where n now is positive integer.
So, f(x)=f(-n)=-f(n)=-kn = k(-n) = kx
So for all integers, f(x) = kx
Next, let x be a rational number: x = p/q where p,q are integers and gcd(p,q)=1.
Then
kp=f(p)=f(q(p/q))=qf(p/q)
so that f(p/q)=k p/q
Hence for rational x also, we have f(x)=kx.
Finally, if x is an irrational point, then we use the continuity of f to prove that f(x)=kx.
More specifically, consider an irrational number x₀. Consider a sequence of rational numbers {x_i} whose limit is x₀. (Such a sequence can always be found since there are infinitely many rationals in the vicinity of any irrational number).
Consider the sequence of the corresponding function values {f(x_i)}. Since all the x_i's are rational, we have
{f(x_i}={kx_i} = k{x_i} → kx₀ ------- (2)
Since the function is continuous, lim of f(x) as x→x₀ is same as f(x₀). Accordingly, for an irrational number x₀,
f(x₀)=lim_{x→x₀} f(x)= k x₀ (from (2))

Hence, for all real x, we get f(x) =k x.